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A person's savings for 3 years are in an arithmetic progression that in the 3 years he has saved cs 24,000 and in the first year he saves more than he saved in the second year

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Answer to a math question A person's savings for 3 years are in an arithmetic progression that in the 3 years he has saved cs 24,000 and in the first year he saves more than he saved in the second year

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Murray
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92 Answers
Let's denote the amount saved in the first year as a , the common difference as d , and the total amount saved in 3 years as 24,000 . Since the savings are in an arithmetic progression, we have the following equations:

1) Amount saved in the 1st year: a
2) Amount saved in the 2nd year: a + d
3) Amount saved in the 3rd year: a + 2d

From the given information, we have the total amount saved in 3 years:
a + (a + d) + (a + 2d) = 24,000

Simplify the equation:
3a + 3d = 24,000

Divide by 3:
a + d = 8,000

Since the amount saved in the 1st year is more than the 2nd year, we have:
a > a + d

Simplify this inequality:
a > 8,000

Therefore, the person's savings for the 3 years are a + (a+d) + (a+2d) = 2a + 3d = a + a + 8,000 = 2a + 8,000 and the amount saved in the first year is greater than 8,000 .

\textbf{Answer:} The person's savings for 3 years are 2a + 8,000 , where a > 8,000 .

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